If Current Through a Resistor Is Doubled, How Does Electrical Power Change?

Correct Answer: (D) 4P

Understanding the Relationship Between Current and Electrical Power

This question examines how the electrical power dissipated in a resistor changes when the current flowing through it is doubled. The resistance R is stated to be constant, so the most useful expression for electrical power is the one that relates power directly to current and resistance.

For a resistor, the electrical power dissipated as heat is given by:

P = I²R

This equation shows that when resistance remains constant, power is proportional to the square of the current. Therefore, any change in current produces a squared effect on the power dissipated by the resistor.

If the current becomes twice its original value, the power does not simply become twice as large. Because the current is squared in the power equation, the new power becomes four times the original power.

Deriving the Electrical Power Formula Using Ohm’s Law

The general formula for electrical power is:

P = VI

According to Ohm’s law:

V = IR

Substituting V = IR into the power formula gives:

P = (IR)I

Therefore:

P = I²R

This form of the electrical power equation is particularly useful when the current changes while the resistance is treated as constant.

Initial Power Dissipated Across the Resistor

Let the initial current flowing through the resistor be I. The resistance of the resistor is R and remains constant.

The initial electrical power is therefore:

P = I²R

This represents the original rate at which electrical energy is converted into heat in the resistor.

Power When the Current Is Doubled

If the initial current is I, doubling the current gives:

I’ = 2I

Let the new power dissipated by the resistor be P’. Since the resistance remains constant:

P’ = (I’)²R

Substituting I’ = 2I:

P’ = (2I)²R

Squaring the factor 2 gives:

P’ = 4I²R

Since the original power is:

P = I²R

we obtain:

P’ = 4P

Therefore, when the current through a resistor of constant resistance is doubled, the power dissipated becomes four times its original value.

Why the Power Becomes Four Times

The key point is that electrical power is proportional to the square of the current when resistance remains constant:

P ∝ I²

Therefore, if the current changes by a factor of 2, the power changes by the square of that factor:

New power factor = 2² = 4

This means that doubling the current increases the power dissipation by a factor of four. Similarly, tripling the current would increase the power by a factor of nine, provided the resistance remains constant.

Ratio Method for Solving the Question Quickly

The same result can be obtained by comparing the initial and final powers.

For the initial current:

P = I²R

For the doubled current:

P’ = (2I)²R

Taking the ratio:

P’/P = [(2I)²R]/[I²R]

The resistance R and I² cancel:

P’/P = 4

Therefore:

P’ = 4P

This ratio method is an efficient way to solve questions involving proportional changes in electrical quantities.

Connection Between Electrical Power and Joule Heating

When electric current flows through a resistor, electrical energy is converted into thermal energy because of collisions between charge carriers and the atoms of the conductor. The rate at which this energy is converted into heat is the electrical power dissipated by the resistor.

According to Joule’s law of heating, the heat produced in a resistor during a time interval t is:

H = I²Rt

Dividing both sides by time gives the rate of heat production:

H/t = I²R

Since power is energy transferred per unit time:

P = I²R

Thus, doubling the current causes four times as much electrical energy to be converted into heat per unit time.

Detailed Analysis of Each Option

Option (A): P/2

Option (A) is incorrect. Doubling the current does not reduce the power to half its original value. For a constant resistance, power is directly proportional to the square of the current. Therefore, an increase in current causes the power to increase rather than decrease.

If the current changes from I to 2I, the power changes from I²R to 4I²R. Hence, the new power cannot be P/2.

Option (B): P

Option (B) is incorrect. The power would remain equal to P only if the quantities determining the power remained unchanged. In this question, the current is doubled, so the power must change.

Since power varies as I² for a constant resistance, doubling the current significantly increases the power dissipation.

Option (C): 2P

Option (C) is incorrect. This option would be obtained by assuming that power is directly proportional to current alone. However, when resistance is constant, the appropriate relation is P = I²R.

Because the current is squared, doubling I produces a factor of 2², which equals 4. Therefore, the new power is not 2P.

Option (D): 4P

Option (D) is correct. The initial power is P = I²R. When the current is doubled, the new current becomes 2I.

Therefore:

P’ = (2I)²R = 4I²R = 4P

Hence, the power dissipated across the resistor becomes four times the original power.

Important Note About the Constant Voltage Statement

The question states that both the resistance R and the voltage V are constant and then asks what happens if the current is doubled. For an ideal ohmic resistor, Ohm’s law gives I = V/R. Therefore, if both V and R remain strictly constant, the current cannot independently double.

The intended interpretation of the question is that the current is changed from I to 2I while the resistance remains constant, and the change in power is determined using P = I²R. Under this intended interpretation, the expected answer is 4P.

Physically, to double the current through the same constant resistance, the applied voltage would also need to double according to Ohm’s law. The mathematical result for the new power would still be four times the original value.

Alternative Verification Using Voltage

According to Ohm’s law:

V = IR

If resistance remains constant and the current is doubled, the required voltage becomes:

V’ = (2I)R = 2V

The power can also be expressed as:

P = V²/R

Therefore, when the voltage becomes 2V:

P’ = (2V)²/R

P’ = 4V²/R

Since V²/R = P:

P’ = 4P

This confirms the result obtained from the relation P = I²R.

Final Answer

For a resistor of constant resistance:

P = I²R

When the current is doubled:

I’ = 2I

Therefore:

P’ = (2I)²R

P’ = 4I²R

P’ = 4P

Therefore, the correct answer is (D) 4P.